Here is the example I encountered :
A matrix $\ M\ $ $(5\times 5)$ is given and its minimal polynomial is determined to be $(x-2)^3.$ So considering the two possible sets of elementary divisors $$\{ (x-2)^3,(x-2)^2\}\ \ and\ \ \{(x-2)^3,(x-2),(x-2)\}$$ we get two possible Jordan Canonical forms of the matrix , namely $J_1$ and $J_2$ respectively. So $J_1$ has $2$ and $J_2$ has $3$ Jordan Blocks.
Now we are to determine the exact one from these two. From the original matrix $M,$ we determined the Eigen vectors and $2$ eigen vectors were linearly independent. So the result is that $J_1$ is the one .So, to determine the exact one out of all possibilities , we needed two information -
$1)$ the minimal polynomial ,together with $2)$ the number of linearly independent eigen vectors .
Now this was a question-answer book so not much theoretical explanations are given . From the given result , I assume the number of linearly independent eigen vectors -which is $2$ in this case - decided $J_1$ to be the exact one because it has $2$ Jordan Blocks. So the equation
"Number of linearly independent eigen vectors$=$Number of Jordan Blocks"
must be true for this selection to be correct .
Now this equation is not proved in this book or the text book I have read says nothing of this sort .
So, that is my question here : How to prove the equation "Number of linearly independent eigen vectors$=$Number of Jordan Blocks" $?$
Let $A$ be a matrix reduced in Jordan form. Notice that for every Jordan block relative to the eigenvalue $\alpha$ there is an eigenvector with respect to $\alpha$ and these vectors ar independent. To see this just look at the first column of a generic block. So $\#$Jordan blocks$\le\#$independent eigenvectors. To prove the other inequality, consider the matrix $A-\alpha I$ for every eigenvalue $\alpha$ and calculate its rank.